Pete's Blog

The following are the first ten terms of a sequence:

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, . . .

Same idea as the other sequences. Describe the pattern of the sequence. If you use a search engine, please don't submit it in the comments.

If it hasn't been solved by Wednesday, I'll give a hint. It's pretty easy, so you probably won't need a hint. Also, if you ask a polar question (i.e., a yes/no question), I will probably answer it if it relates to this sequence.

Update: Hint: It's related to powers of 2.

I'll post the solution on Thursday. (Since I forgot to put the hint up for last week's sequence until today, I won't put the solution up for it until Wednesday of this week.)

Update: The sequence is the highest exponent for which 2 to the exponent of the entry divides the index. E.g., 2¹=2 divides 3 but 2²=4 does not, so a(3)=1. E.g., 2³=8 divides 8 but 2⁴=16 does not, so a(8)=3. E.g., 2⁰=1 divides 7 but 2 does not, so a(7)=0. E.g., 2¹=2 divides 10 but 2²=4 does not, so a(10)=1. It was not solved.

Update: Last week's second sequence (1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, . . .) was the list of smallest integers with exactly the index's number of proper divisors, that is, a(n) was the smallest positive integer with exactly n divisors. E.g., 1, 2, 3, 4, 6, 9, 12, 18, 36 all divide 36, and they are the only divisors of 36. Furthermore, no positive integer less than 36 has exactly nine divisors. So a(9)=36. It was not solved. As an extra exercise, prove that for each i=1, 2, 3, . . . there exists an integer with exactly i divisors.

The following is a list of sequences (of varying difficulty, where the difficulty is reported based on the number of terms listed, on my subjective assessment) for the following seven weeks, each with a varying number of terms listed:

1. 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, . . .
   
2. 5, 7, 13, 19, 31, 43, 61, 73, 103, 109, 139, 151, 181, 193, 199, 229, 241, 271, 283, 313, 349, 421, 433, 463, 523, 571, 601, 619, 643, 661, 811, 823, 829, 859, 883, 1021, 1033, 1051, 1063, 1093, 1153, 1231, 1279, 1291, 1303, 1321, 1429, 1453, 1483, 1489, . . . (Update: Hint: Suppose p and q are primes. Which for which values of p and q is p + q in the sequence?)
   
3. 4, 8, 9, 12, 16, 18, 20, 24, 25, 27, . . . (Update: Hint: Look at the prime factors.)
   
4. 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, . . . (Update: Hint: (Upon further thinking, I should have given more terms to indicate that this sequence was tough.) It's something to do with 2^a(n).)
   
5. 1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 22, 11, . . . (Update: Hint: First, notice that the greatest common divisor for successive terms beyond the first (i.e., gcd(a(i), a(i + 1))>1 for each i є {2, 3, 4, . . .}. Second, this sequence contains every positive integer.)
   
6. 1, 4, 6, 7, 8, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 23, 25, 27, 29, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 70, 71, 72, 73, 74, . . . (Update: Hint: Notice that i is in the sequence if and only if a(i) is odd.)
   
7. 5, 53, 157, 173, 211, 257, 263, 373, 563, 593, 607, 653, 733, 947, 977, 1103, 1123, 1187, 1223, 1367, . . . (Update: Hint: This sequence is also a subset of the primes. It's somewhat similar to the sequence five above. For each positive integer i, a(i) is related to two primes, p and q, and more specifically the sum p + q. But, this sequence is dissimilar from the one above in that this one's terms, while all are directly related to p + q they're not necessarily of the form p + q. The terms are of the form p + q with something more done to them.)
   

Same idea as the other sequences. Describe the pattern of the sequence. If you use a search engine, please don't submit it in the comments.

Within eight weeks from now, for any of the sequences that are still left unsolved I'll post a hint, and then on June 15^th, with that week's sequence, post the solutions.